Given a function 'g' continous everywhere such that `int _(0) ^(1) g (t ) dt =2 and g (1)=5.` If `f (x ) =1/2 int _(0) ^(x) (x-t) ^(2)g (t)dt,` then the vlaue of `f'''(1)-f''(1)` is:

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the value of \( f'''(1) - f''(1) \) given the function \( f(x) \) defined as:

\[ f(x) = \frac{1}{2} \int_0^x (x-t)^2 g(t) \, dt \]

where \( g(t) \) is a continuous function satisfying \( \int_0^1 g(t) \, dt = 2 \) and \( g(1) = 5 \).

### Step 1: Determine the function \( g(t) \)

We know that:

\[ \int_0^1 g(t) \, dt = 2 \]

and \( g(1) = 5 \). We can assume a form for \( g(t) \). Let's try \( g(t) = a t^b \).

From \( g(1) = 5 \), we have:

\[ a \cdot 1^b = 5 \implies a = 5 \]

Now substituting into the integral:

\[ \int_0^1 5 t^b \, dt = 2 \]

Calculating the integral:

\[ 5 \cdot \frac{t^{b+1}}{b+1} \bigg|_0^1 = 5 \cdot \frac{1}{b+1} = 2 \implies \frac{5}{b+1} = 2 \implies 5 = 2(b+1) \implies 5 = 2b + 2 \implies 2b = 3 \implies b = \frac{3}{2} \]

Thus, we have:

\[ g(t) = 5 t^{\frac{3}{2}} \]

### Step 2: Substitute \( g(t) \) into \( f(x) \)

Now substituting \( g(t) \) back into \( f(x) \):

\[ f(x) = \frac{1}{2} \int_0^x (x-t)^2 \cdot 5 t^{\frac{3}{2}} \, dt = \frac{5}{2} \int_0^x (x-t)^2 t^{\frac{3}{2}} \, dt \]

### Step 3: Expand \( (x-t)^2 \)

Using the binomial expansion:

\[ (x-t)^2 = x^2 - 2xt + t^2 \]

Thus, we can write:

\[ f(x) = \frac{5}{2} \int_0^x (x^2 - 2xt + t^2) t^{\frac{3}{2}} \, dt \]

### Step 4: Break down the integral

Now we can separate the integral:

\[ f(x) = \frac{5}{2} \left( x^2 \int_0^x t^{\frac{3}{2}} \, dt - 2x \int_0^x t^{\frac{5}{2}} \, dt + \int_0^x t^{\frac{7}{2}} \, dt \right) \]

Calculating each integral:

1. \( \int_0^x t^{\frac{3}{2}} \, dt = \frac{2}{5} x^{\frac{5}{2}} \)
2. \( \int_0^x t^{\frac{5}{2}} \, dt = \frac{2}{7} x^{\frac{7}{2}} \)
3. \( \int_0^x t^{\frac{7}{2}} \, dt = \frac{2}{9} x^{\frac{9}{2}} \)

Substituting back:

\[ f(x) = \frac{5}{2} \left( x^2 \cdot \frac{2}{5} x^{\frac{5}{2}} - 2x \cdot \frac{2}{7} x^{\frac{7}{2}} + \frac{2}{9} x^{\frac{9}{2}} \right) \]

This simplifies to:

\[ f(x) = \frac{5}{2} \left( \frac{2}{5} x^{\frac{9}{2}} - \frac{4}{7} x^{\frac{9}{2}} + \frac{2}{9} x^{\frac{9}{2}} \right) \]

### Step 5: Combine the terms

Combining the coefficients:

\[ f(x) = \frac{5}{2} x^{\frac{9}{2}} \left( \frac{2}{5} - \frac{4}{7} + \frac{2}{9} \right) \]

Calculating the common denominator (which is 315):

\[ \frac{2}{5} = \frac{126}{315}, \quad \frac{4}{7} = \frac{180}{315}, \quad \frac{2}{9} = \frac{70}{315} \]

So:

\[ \frac{126 - 180 + 70}{315} = \frac{16}{315} \]

Thus:

\[ f(x) = \frac{5}{2} \cdot \frac{16}{315} x^{\frac{9}{2}} = \frac{40}{315} x^{\frac{9}{2}} = \frac{8}{63} x^{\frac{9}{2}} \]

### Step 6: Differentiate \( f(x) \)

Now we differentiate \( f(x) \):

1. **First derivative**:
\[ f'(x) = \frac{8}{63} \cdot \frac{9}{2} x^{\frac{7}{2}} = \frac{36}{63} x^{\frac{7}{2}} = \frac{4}{7} x^{\frac{7}{2}} \]

2. **Second derivative**:
\[ f''(x) = \frac{4}{7} \cdot \frac{7}{2} x^{\frac{5}{2}} = 2 x^{\frac{5}{2}} \]

3. **Third derivative**:
\[ f'''(x) = 2 \cdot \frac{5}{2} x^{\frac{3}{2}} = 5 x^{\frac{3}{2}} \]

### Step 7: Evaluate at \( x = 1 \)

Now we evaluate \( f'''(1) \) and \( f''(1) \):

\[ f'''(1) = 5 \cdot 1^{\frac{3}{2}} = 5 \]

\[ f''(1) = 2 \cdot 1^{\frac{5}{2}} = 2 \]

### Step 8: Calculate \( f'''(1) - f''(1) \)

Finally, we find:

\[ f'''(1) - f''(1) = 5 - 2 = 3 \]

### Final Answer

Thus, the value of \( f'''(1) - f''(1) \) is:

\[ \boxed{3} \]